\(\int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\) [327]

Optimal result

Integrand size = 29, antiderivative size = 101 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {5 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{4 d}-\frac {a \cot (c+d x)}{4 d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)}}{2 d} \] Output:

5/4*a^(1/2)*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/d-1/4*a*cot 
(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-1/2*cot(d*x+c)*csc(d*x+c)*(a+a*sin(d*x+c) 
)^(1/2)/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(101)=202\).

Time = 0.97 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.47 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\csc ^7\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+6 \cos \left (\frac {3}{2} (c+d x)\right )-5 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+5 \cos (2 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+5 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-5 \cos (2 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )+6 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{4 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^2} \] Input:

Integrate[Cot[c + d*x]^2*Csc[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

-1/4*(Csc[(c + d*x)/2]^7*Sqrt[a*(1 + Sin[c + d*x])]*(2*Cos[(c + d*x)/2] + 
6*Cos[(3*(c + d*x))/2] - 5*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 
5*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 5*Log[1 
- Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 5*Cos[2*(c + d*x)]*Log[1 - Cos[(c 
 + d*x)/2] + Sin[(c + d*x)/2]] - 2*Sin[(c + d*x)/2] + 6*Sin[(3*(c + d*x))/ 
2]))/(d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^2 
)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3353, 3042, 3454, 27, 3042, 3459, 3042, 3252, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^2 \sqrt {a \sin (c+d x)+a}}{\sin (c+d x)^3}dx\)

\(\Big \downarrow \) 3353

\(\displaystyle \frac {\int \csc ^3(c+d x) (a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}dx}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}{\sin (c+d x)^3}dx}{a^2}\)

\(\Big \downarrow \) 3454

\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{2} \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} \left (a^2-3 a^2 \sin (c+d x)\right )dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} \left (a^2-3 a^2 \sin (c+d x)\right )dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \frac {\sqrt {\sin (c+d x) a+a} \left (a^2-3 a^2 \sin (c+d x)\right )}{\sin (c+d x)^2}dx-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 3459

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {5}{2} a^2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {a^3 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (-\frac {5}{2} a^2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {a^3 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {1}{4} \left (\frac {5 a^3 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {a^3 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{4} \left (\frac {5 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {a^3 \cot (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )-\frac {a^2 \cot (c+d x) \csc (c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}}{a^2}\)

Input:

Int[Cot[c + d*x]^2*Csc[c + d*x]*Sqrt[a + a*Sin[c + d*x]],x]
 

Output:

(-1/2*(a^2*Cot[c + d*x]*Csc[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d + ((5*a^( 
5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (a^3*Co 
t[c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/4)/a^2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, 
 e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3353

Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/b^2   Int[(d*Sin[e 
 + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /; Fre 
eQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] ||  !IGtQ[ 
n, 0])
 

rule 3454

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x 
] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 
-1]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (5 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right )^{2} a^{2}+3 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} \sqrt {a}-5 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {3}{2}}\right )}{4 a^{\frac {3}{2}} \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(126\)

Input:

int(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE 
)
 

Output:

1/4*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(3/2)*(5*arctanh((-a*(sin(d 
*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+3*(-a*(sin(d*x+c)-1))^(3/2)*a^(1 
/2)-5*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2))/sin(d*x+c)^2/cos(d*x+c)/(a+a*sin( 
d*x+c))^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (85) = 170\).

Time = 0.09 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.16 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {5 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} + {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + 2 \, \cos \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) - d\right )}} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="fri 
cas")
 

Output:

1/16*(5*(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + 
c) - cos(d*x + c) - 1)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 
+ 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3 
)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 
+ 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 
 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) + 4*(3*cos(d*x + 
 c)^2 + (3*cos(d*x + c) + 1)*sin(d*x + c) + 2*cos(d*x + c) - 1)*sqrt(a*sin 
(d*x + c) + a))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2 - d*cos(d*x + c) + (d 
*cos(d*x + c)^2 - d)*sin(d*x + c) - d)
 

Sympy [F]

\[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )} \cot ^{2}{\left (c + d x \right )} \csc {\left (c + d x \right )}\, dx \] Input:

integrate(cot(d*x+c)**2*csc(d*x+c)*(a+a*sin(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))*cot(c + d*x)**2*csc(c + d*x), x)
 

Maxima [F]

\[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cot \left (d x + c\right )^{2} \csc \left (d x + c\right ) \,d x } \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="max 
ima")
 

Output:

integrate(sqrt(a*sin(d*x + c) + a)*cot(d*x + c)^2*csc(d*x + c), x)
 

Giac [A] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.53 \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (5 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {4 \, {\left (6 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}\right )} \sqrt {a}}{16 \, d} \] Input:

integrate(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x, algorithm="gia 
c")
 

Output:

1/16*sqrt(2)*(5*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2 
*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 
1/2*d*x + 1/2*c)) - 4*(6*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 
 1/2*d*x + 1/2*c)^3 - 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 
1/2*d*x + 1/2*c))/(2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2)*sqrt(a)/d
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int -\frac {\left ({\sin \left (c+d\,x\right )}^2-1\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\sin \left (c+d\,x\right )}^3} \,d x \] Input:

int((cot(c + d*x)^2*(a + a*sin(c + d*x))^(1/2))/sin(c + d*x),x)
 

Output:

int(-((sin(c + d*x)^2 - 1)*(a + a*sin(c + d*x))^(1/2))/sin(c + d*x)^3, x)
 

Reduce [F]

\[ \int \cot ^2(c+d x) \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\sqrt {a}\, \left (\int \sqrt {\sin \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2} \csc \left (d x +c \right )d x \right ) \] Input:

int(cot(d*x+c)^2*csc(d*x+c)*(a+a*sin(d*x+c))^(1/2),x)
 

Output:

sqrt(a)*int(sqrt(sin(c + d*x) + 1)*cot(c + d*x)**2*csc(c + d*x),x)